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I don't know how to answer this question:

Is there a $10 \times 10$ matrix $A$ such that $$M_{10}(\mathbb{F})=\text{span}\{I,A,A^2,\ldots, A^{100}\}\textrm{,}$$ where $M_{10}(\mathbb{F})$ is the vector space of $10 \times 10$ matrices over $\mathbb{F}$?

I think there is a connection to Jordan normal form.

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There are many reasons why this is not possible. Certainly citing the Cayley-Hamilton theorem comes to mind, from which it follows that already $A^{10}$ is certainly linearly dependent on the previous powers. But there is also a very basic argument: if there were such a matrix $A$, then all $10\times10$ matrices, being polynomials in$~A$, would commute among each other. Which they don't.

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  • $\begingroup$ @Marc Yes, this last argument is especially slick! $\endgroup$ – Travis Willse Sep 20 '14 at 18:26
  • $\begingroup$ +1. Thinking about what commutation relations can be a simple but powerful for thinking about such things. $\endgroup$ – asmeurer Sep 20 '14 at 19:10
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    $\begingroup$ @asmeurer: It is really a lot like saying a non-Abelian group cannot be cyclic, which is a very obvious thing to realise. Somehow the ring theoretic counterpart looks a bit less obvious. $\endgroup$ – Marc van Leeuwen Sep 20 '14 at 21:45
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    $\begingroup$ @Benjamin: What makes you think the argument is more clear for those two fields than for any other field? It must be your own familiarity with those fields, because my argument does not in fact use any specific property of the field. Commutativity of scalars is all that matters (and without it the question would not really makes sense). $\endgroup$ – Marc van Leeuwen Apr 30 '15 at 3:39
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    $\begingroup$ I think being "a bit less obvious" is because it simply involves one more operation than the group-theoretic version and polynomials are more distracting than monomials (powers of a generating element). $\endgroup$ – Vandermonde Oct 18 '15 at 2:23
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Any $n \times n$ matrix $A$ satisfies $p_A(A) = 0$, where $p_A$ is the (degree n) characteristic polynomial of $A$; in particular, $A^n$ can be written as a linear combination of $I, A, A^2, \ldots, A^{n - 1}$, and thus by induction so can any higher power of $A$. So, the span over $\mathbb{F}$ of $\{I, A, A^2, \ldots\}$ has dimension at most $n$, which (for $n > 1$) is smaller than the $n^2 = \dim_{\mathbb{F}} M_n(\mathbb{F})$ asked.

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For reasons explained by others no single one matrix can provide a basis for all the other matrices.

However The Jordan normal form is interesting to consider if we want to investigate the requirements for being able to express one matrix $A$ in terms of another $B$. They obviously must have at least pairwise same dimensionality of generalized eigenspaces because the block structure is preserved by any polynomial (or any other function also). Then we can try and find pairwise equivalence relations for the generalized eigenspaces. $T_iA_iT_i^{-1} = B_i$ how this would translate to requirements for a polynomial.. I don't know. But it is an interesting question!

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